Work Energy Theorem

8 MCQs3 revision cards9-step worked example

Source: NCERT Work, Energy and PowerPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The work-energy theorem is one sentence that connects force, displacement, and motion into a single bookkeeping equation — and NEET exploits the fact that students botch the bookkeeping.

The theorem (NCERT Class 11 Physics Chapter 5, page 3): The net work done by all forces acting on a particle equals the change in its kinetic energy.

W_net = ΔK = ½ m v² − ½ m v₀²

"All forces" is the trap. W_net is not the work of gravity alone, not the work of the applied force alone — it is the algebraic sum of work done by every force: gravity, normal, friction, applied, spring, tension. Miss one force, lose four marks.

Where students lose marks:

Normal force on a horizontal surface. The normal force is perpendicular to displacement → cos 90° = 0 → work = zero. A common distractor offers W = N × s as if the normal force does positive work. It does not.
Forgetting that W_net includes friction. On a rough surface the theorem reads: W_applied + W_gravity + W_friction = ΔK. If you write mgh = ½mv² on a rough incline, you have silently dropped the friction term. The theorem still holds — you just used it incorrectly by omitting a force.
Confusing displacement with velocity in power problems that invoke WET. The theorem links work (force × displacement) to kinetic energy change (involving speed squared). When a follow-up asks for power, remember P = F · v, not F · x.

Bridge to NEET: Questions from 2021–2025 test WET through hydroelectric turbines (with efficiency losses), rough-vs-smooth incline comparisons, and bullet-into-block deceleration problems. In every case the template is identical: identify all forces, compute each work term, equate the sum to ΔK.

Watch-out: If the problem states "constant speed," then ΔK = 0 and W_net = 0. That does not mean no force does work — it means the positive and negative work terms cancel exactly.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

The work-energy theorem states that the net work done on a particle equals:

MCQ 2Easy RecallPractice

A block slides on a horizontal surface. The work done by the normal reaction force on the block is:

MCQ 3Easy RecallPractice

In the work-energy theorem W_net = ΔK, the term W_net refers to:

MCQ 4Direct ApplicationPractice

A 2.0 kg block initially at rest is pushed by a constant horizontal force of 10 N across a frictionless surface for 5.0 m. Using the work-energy theorem, the final speed of the block is:

MCQ 5Direct ApplicationPractice

A body moving at speed v is brought to rest by a constant braking force over a distance d. If the initial speed is doubled to 2v, the stopping distance under the same braking force becomes:

MCQ 6Direct ApplicationPractice

A block of mass 4.0 kg slides down a rough incline (height 5.0 m). It reaches the bottom with a speed of 6.0 m/s. Taking g = 10 m/s², the work done by friction during the descent is:

MCQ 7CalculationPractice

A bullet of mass 20 g moving at 200 m/s enters a wooden block and penetrates 25 cm before stopping. A second identical bullet moving at 100 m/s enters an identical block. The penetration depth of the second bullet is:

MCQ 8CalculationPractice

A 2.0 kg block at rest on a frictionless surface is acted on by a position-dependent net force F_net(x) = (8.0x) N from x = 0 to x = 3.0 m. Using the work-energy theorem, the speed of the block at x = 3.0 m is:

Quick recall before you leave

Worked Example

Pattern: Hydroelectric turbine power with efficiency loss (anchored to PYQ pattern NEET pattern: hydroelectric turbine power, source: PYQ.2021.P3.Q035).

1
Given
- Height: h = 60 m - Flow rate: dm/dt = 15 kg/s - Energy loss: 10% (efficiency η = 0.90) - g = 10 m/s² (exact, problem-defined)
2
Required
Useful power output P_useful.
3
Concept
By the work-energy theorem, the gravitational PE of falling water is converted to KE at the turbine. The rate of energy conversion is the power input. The turbine extracts a fraction η of this power.
4
Formula
P_input = (dm/dt) × g × h P_useful = η × P_input = η × (dm/dt) × g × h
5
Substitution
P_useful = 0.90 × 15 × 10 × 60
6
Calculation
P_useful = 0.90 × 15 × 10 × 60 = 0.90 × 9000 = 8100 W = 8.1 kW **Note on exact values:** g = 10 m/s² is an exact value specified by the problem. The integers 15, 60, and the efficiency factor 0.90 are all exact as given. Significant-figure limits are not the binding constraint here — the answer is exact within the problem's defined quantities.
7
Final answer
P_useful = 8.1 kW
8
Common trap
The high-frequency trap here is forgetting the efficiency factor (trap: trap: power efficiency factor drop). Without the 0.90 multiplier, you get P = 9.0 kW — which is a typical distractor in NEET. Always check whether the problem states an efficiency, loss percentage, or "10% is wasted."
9
Similar NEET-style question
A pump lifts water from a well of depth 40 m at a rate of 20 kg/s. If the pump is 80% efficient, what is the power consumed by the pump? (Answer: P_consumed = (dm/dt × g × h)/η = 20 × 10 × 40 / 0.80 = 10 000 W = 10 kW. Note the inversion: here you solve for input power given useful work, so divide by η.) ---

Before solving, remember these

Law

Work-energy theorem

The work done by the net force on a particle equals the change in its kinetic energy: W_net = ΔK = K_f - K_i = ½ m (v² - v₀²). Holds for both constant and variable forces and for one or more dimensions.

-- NCERT Class 11 Physics, Ch. 5, p. 3

Formulas

10 formulas — click to collapse

Elastic collision — 1D final velocities

v_{1} = \frac{( m _{1} - m _{2} ) u _{1} + 2 m _{2} u _{2}}{m _{1} + m _{2}}, v_{2} = \frac{( m _{2} - m _{1} ) u _{2} + 2 m _{1} u _{1}}{m _{1} + m _{2}}

Final velocities of two bodies after an elastic head-on (1D) collision — momentum AND kinetic energy are both conserved. Special cases: equal masses exchange velocities; very heavy m2 at rest reflects m1 with reversed velocity.

Symbol	Quantity	SI Unit
m1, m2	Masses of the two bodies	kg
u1, u2	Initial velocities (signed, before collision)	m/s
v1, v2	Final velocities (signed, after collision)	m/s

Valid when

Collision is ELASTIC (kinetic energy conserved)
Head-on (1D) — for 2D collisions decompose along/perpendicular to line of impact

Do NOT use when

Inelastic collision (KE not conserved; use momentum-only + restitution)
Bodies stick together (perfectly inelastic case has its own formula)

Gravitational potential energy (near Earth)

U = m g h

Potential energy of a body of mass m at height h above the chosen reference level, in a region where g is approximately constant.

Symbol	Quantity	SI Unit
U	Gravitational PE	J
m	Mass	kg
g	Gravitational acceleration	m/s^2
h	Height above reference level	m

Valid when

Region small enough that g is uniform (typically near Earth's surface)
Reference level is freely chosen — only DIFFERENCES in U have physical meaning

Do NOT use when

Large altitude changes (use U = -GMm/r general form)
Below the reference level — h is signed

Kinetic energy

K = \frac{1}{2} m v^{2}

Energy a body possesses by virtue of its motion. Always non-negative. Frame-dependent through v.

Symbol	Quantity	SI Unit
K	Kinetic energy	J
m	Mass	kg
v	Speed	m/s

Valid when

Non-relativistic speeds (v << c)
Translational motion only (rotational KE = (1/2) I omega^2 separately)

Conservation of mechanical energy

K_{i} + U_{i} = K_{f} + U_{f}

If only conservative forces do work on a system, the total mechanical energy E = K + U is constant in time.

Symbol	Quantity	SI Unit
K_i, K_f	Initial, final kinetic energy	J
U_i, U_f	Initial, final potential energy	J

Valid when

Only CONSERVATIVE forces do work (gravity, spring, electrostatic — not friction or drag)
Closed system; no energy exchange with surroundings

Do NOT use when

Friction, drag, or other non-conservative forces do work
External forces add energy to the system

Perfectly inelastic 1D collision — common final velocity and KE loss

v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}, Δ K = - \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} (u_{1} - u_{2})^{2}

When two bodies stick together after a 1D collision, the common velocity is given by momentum conservation. The KE lost is converted to internal energy (heat, deformation).

Symbol	Quantity	SI Unit
v	Common final velocity	m/s
m1, m2	Masses	kg
u1, u2	Initial velocities	m/s
Delta_K	Change in kinetic energy	J

Valid when

Bodies stick together immediately after collision (perfectly inelastic)
Net external force = 0 during the brief collision (momentum conservation)

Instantaneous power

P = \frac{d W}{d t} = F \cdot v

Instantaneous power is the time rate of doing work; equivalently, the dot product of the force and velocity. Average power over an interval is W/t.

Symbol	Quantity	SI Unit
P	Power (instantaneous)	W
F	Force	N
v	Velocity	m/s

Valid when

F.v form when force and velocity are both known at the instant

Spring potential energy (Hooke's law regime)

U = \frac{1}{2} k x^{2}

Elastic potential energy stored in an ideal spring of force constant k that has been displaced by x from its natural length. Independent of the sign of x (compression or extension).

Symbol	Quantity	SI Unit
U	Elastic PE	J
k	Spring constant (force per unit displacement)	N/m
x	Displacement from natural length	m

Valid when

Spring obeys Hooke's law (F = -k*x) over the displacement range
No internal damping / hysteresis assumed

Do NOT use when

Beyond elastic limit (Hooke's law fails)
Real springs with finite damping (some elastic PE goes to internal energy)

Work done by a constant force

W = F \cdot s = F s cos θ

The work done by a constant force F on an object that undergoes a displacement s is the dot product F.s. Equivalently, W = (magnitude of F) * (magnitude of s) * cos(angle between them). Work is a scalar but has a sign.

Symbol	Quantity	SI Unit
W	Work (scalar, signed)	J
F	Constant force (vector)	N
s	Displacement (vector)	m
theta	Angle between F and s	rad/deg

Valid when

Force is CONSTANT in magnitude and direction over the displacement
Use the COMPONENT of force along the displacement, not magnitude alone

Do NOT use when

Force varies with position (use the variable-force integral)
Multiple forces act — apply this to each one separately or use net force

Work-energy theorem

W_{net} = Δ K = \frac{1}{2} m (v^{2} - v_{0}^{2})

The net work done by all forces on a particle equals the change in its kinetic energy. Holds for both constant and variable forces, in 1D and higher dimensions.

Symbol	Quantity	SI Unit
W_net	Net work done by all forces	J
Delta_K	Change in kinetic energy	J
m	Mass of particle	kg
v, v0	Final, initial speed	m/s

Valid when

W_net is the NET (vector-sum) work, not work of any one force
Particle (point-mass) idealisation; for extended bodies handle internal energy separately

Work done by a variable force

W = \int_{x_{i}}^{x_{f}} F (x) d x

When a force varies along the path, the work done is the line integral of force over displacement. In one dimension this is the area under the F-vs-x curve between the start and end positions.

Symbol	Quantity	SI Unit
W	Work done	J
F(x)	Force as a function of position	N
dx	Infinitesimal displacement	m
x_i, x_f	Initial and final positions	m

Valid when

Force may depend on position (e.g. spring force F = -k*x)
Generalises to W = integral F.dr in higher dimensions

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

12 items — click to collapse

Trap

Linear speed-distance ratio under uniform deceleration

Category: Overthinking

Student assumes proportionality of speed to remaining distance under uniform deceleration. In fact, KE drops linearly with distance (v² is the linear quantity, not v): v² = u² - 2as. Speed-vs-distance is a sqrt-curve, not a line.

When it triggers

Question describes a body decelerating through stages with given speed at one stage; asks for distance to stop or speed at another stage.

How to avoid

Always work with v², not v, when uniform deceleration is in play. The work-energy theorem gives the same answer faster: ½ m v² = work done against constant force over distance.

Trap

Incline-with-friction: missing μ cos θ term

Category: Sign Convention

Student writes a = g sin θ for a rough incline (which is the smooth-incline answer); forgets to subtract μ g cos θ.

When it triggers

Question contrasts rough vs smooth inclines, or asks for acceleration on a rough incline.

How to avoid

On a rough incline (block sliding down): a = g(sin θ - μ_k cos θ). On a rough incline (block sliding up): a = -g(sin θ + μ_k cos θ). Smooth case (μ = 0): just g sin θ.

Trap

Inelastic collision: defaulting to elastic

Category: Similar Terms

Student uses the elastic-collision velocity formulas (m1-m2)/(m1+m2) when the question explicitly says 'completely inelastic' (bodies stick).

When it triggers

Question says 'inelastic', 'stick together', 'after collision moves with common velocity'.

How to avoid

Perfectly inelastic 1D: v_common = (m₁ u₁ + m₂ u₂)/(m₁ + m₂). KE is NOT conserved; loss = ½ (m₁ m₂)/(m₁+m₂) × (u₁ - u₂)². Don't use elastic formulas.

Trap

Lift-power: friction term overlooked

Category: Overthinking

Student computes P = Mgv (just lifting against gravity) and ignores the friction-opposing-motion term.

When it triggers

Question describes a lift moving at constant speed with explicit friction force on cable or guides.

How to avoid

At constant speed, net force = 0, so cable tension T = Mg + f_friction. Power = T × v = (Mg + f) × v. Always add friction when stated.

Trap

Power problems: efficiency / loss factor dropped

Category: Overthinking

Student computes ideal power and forgets the (1 - loss_fraction) or efficiency multiplier.

When it triggers

Question gives turbine, motor, or transformer with stated efficiency or loss percentage.

How to avoid

Always read the question for efficiency η or loss%. Useful power P_useful = η × P_input or P_input × (1 - loss). Don't drop the factor even if the rest of the calc is in the unrelated parts of the problem.

Trap

Instantaneous power: x vs v

Category: Similar Terms

Student plugs displacement x into P = F·v formula instead of velocity v.

When it triggers

Question gives displacement x(t) explicitly and a constant force; asks for instantaneous power.

How to avoid

P = F · v where v = dx/dt. Compute v first (differentiate x(t) once), THEN plug into F·v. P is NOT F·x.

Trap

Spring PE linear vs quadratic scaling

Category: Overthinking

Student treats spring PE as proportional to displacement (linear) instead of displacement-squared (quadratic). Common error: 'doubling the stretch doubles the PE'. Actual: doubling the stretch gives 4× the PE.

When it triggers

Question gives U at one stretch and asks for U at another. Distractors include linear-scaling answer (×2 instead of ×4 for double stretch).

How to avoid

U = ½ k x² is QUADRATIC. The PE-to-stretch ratio is the SQUARE of the stretch ratio: if stretch goes from x₁ to x₂, U_new / U_old = (x₂/x₁)².

Trap

Vertical circle: energy conservation alone is insufficient

Category: Overthinking

Student uses ½ m v₀² = m g (2L) (energy to reach top) and forgets the additional v_top² ≥ gL constraint for tension.

When it triggers

Question asks for minimum v₀ at lowest point so the bob can complete a full vertical circle.

How to avoid

TWO constraints: (1) energy: v_top² = v₀² - 4gL; (2) tension at top ≥ 0: v_top² ≥ gL. Combined: v₀² ≥ 5gL. Energy alone gives only v₀² ≥ 4gL which is insufficient.

Mistake

Student applies elastic-collision velocity formulas to an inelastic collision (or vice versa).

Root cause: concept gap

Correction

Elastic: BOTH momentum and kinetic energy conserved -> use the (m1-m2)/(m1+m2) form. Inelastic: ONLY momentum conserved; KE generally lost to heat. Perfectly inelastic: bodies stick together -> common velocity = (m1*u1 + m2*u2)/(m1+m2). Identify which type from the problem before choosing a formula.

Wrong option pattern

Distractor uses elastic-collision formulas for two bodies that the problem says stick together after impact.

Mistake

Student applies K_i + U_i = K_f + U_f in a problem where friction or other non-conservative forces clearly do work.

Root cause: formula misuse

Correction

Mechanical-energy conservation requires that ONLY conservative forces do work. When friction or drag is present, use the work-energy theorem directly: K_f - K_i = W_conservative + W_non-conservative, where W_non-conservative is typically negative (energy goes to heat).

Wrong option pattern

Distractor sets m*g*h = (1/2)*m*v^2 for a block sliding down a rough incline.

Mistake

Student treats gravitational potential energy as having an absolute value, instead of being defined relative to a chosen reference level.

Root cause: concept gap

Correction

PE is defined up to an additive constant. Only DIFFERENCES in PE have physical meaning (they equal the negative work done by the conservative force between two points). Choosing the ground as zero is a convention, not a derivation.

Wrong option pattern

Distractor offers a numerical PE value where two different reference levels would give different correct numbers.

Mistake

Student claims the normal reaction does positive (or negative) work on a block sliding on a horizontal surface.

Root cause: direction ignored

Correction

Work = F . s = F * s * cos(theta). For a block sliding horizontally, the normal force is perpendicular to displacement (theta = 90 deg, cos = 0), so the work done by N is ZERO. Always check the angle between force and displacement before computing work.

Wrong option pattern

Distractor reports W = N * s for a block sliding on a level surface.

Past Year Questions

7 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys. — click to collapse

NEET 2025

A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v 0 as shown in figure. If the string gets slack at some point P making an angle θ from the horizontal, the ratio of the speed v of the bob at point P to its initial speed v 0 is: 1  sinθ 2 1

1 2+3sinθ  

2(sinθ) 2 1 1  1 2  cosθ 2

3 

4  2+3sinθ 2+3sinθ

NTA Answer: Option 1(final)

NEET 2025

There are two inclined surfaces of equal length (L) and same angle of inclination 45° with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on rough surface than on the smooth surface. The coefficient of kinetic friction (µ k) between the object and the rough surface is close to

10.75

20.25

30.40

40.5

NTA Answer: Option 1(final)

NEET 2024Revised key

At any instant of time t, the displacement of any particle is given by 2t – 1 (SI unit) under the influence of force of 5N. The value of instantaneous power is (in SI unit):

110

NTA Answer: Option 1(revised_final)

NEET 2024Revised key

Two bodies A and B of same mass undergo completely inelastic one dimensional collision. The body A moves with velocity v1 while body B is at rest before collision. The velocity of the system after collision is v2. The ratio v1 : v2 is

11 : 2

22 : 1

34 : 1

41 : 4

NTA Answer: Option 2(revised_final)

NEET 2023

The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, potential energy stored in it will be

14 U

28 U

316 U

42 U

NTA Answer: Option 3(final)

NEET 2022

An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms–1. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is : (g = 10 m s–2)

123500

223000

320000

434500

NTA Answer: Option 4(final)

NEET 2021

Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine? m/s2) (g = 10

17.0 kW

210.2 kW

38.1 kW

412.3 kW

NTA Answer: Option 3(final)

How NEET usually asks this

7 recurring patterns from past papers — click to collapse

Two inclines of equal length L and same angle θ (e.g. 45°); one rough (with μ), one smooth. Compare time-of-descent or final velocity at bottom. Smooth: a = g sin θ. Rough: a = g(sin θ - μ cos θ). Use L = ½ a t² or v² = 2aL.

Multi StepMedium

Common distractors

ignores mu cos theta term

Forgetting the friction-along-incline component

Water falls from height h at flow rate dm/dt (kg/s); turbine efficiency or loss factor given; find usable power. Power = dm/dt × g × h × (1 - loss_fraction). Common shape: 60 m, 15 kg/s, 10% loss → P = 15 × 10 × 60 × 0.9 = 8.1 kW.

Multi StepEasy

Common distractors

forgets loss factor

Default to ideal-case formula

Given displacement x(t) and net force F (constant); find instantaneous power P = F · v(t) where v(t) = dx/dt. Common shape: x = 2t-1, F = 5 N → v = 2 m/s, P = 10 W.

Direct ApplicationEasy

Common distractors

uses x instead of v

Default plugging x into the wrong formula

Lift moving up at constant speed v with total mass M; friction force f opposes motion. Power required from cable = (Mg + f) × v. Common shape: M = 2000 kg, v = 1.5 m/s, f given; find motor power.

Multi StepEasy

Common distractors

forgets friction term

Ignoring opposing force

Two bodies of equal mass m, one moving at v₁ and one at rest, undergo completely inelastic 1D collision. Common velocity = v₁/2 (momentum conservation). KE loss = ½ m v₁² × ½ = ¼ m v₁².

Direct ApplicationEasy

Common distractors

no KE loss stated

Treats collision as elastic by default

Spring PE = 1/2 k x^2; given U at one displacement, find U at another. Quadratic scaling: U(x2)/U(x1) = (x2/x1)^2. Common shape: U at 2 cm = U; find U at 8 cm -> 16 U. Distractors test linear (4U) and inverse (U/16) scaling errors.

Direct ApplicationEasy

Common distractors

linear scaling

Thinking PE scales as x not x^2

Bob of mass m on string of length l given horizontal velocity v₀ at lowest point; find condition for string to remain taut throughout the vertical circle. Minimum: v₀² ≥ 5gl. At top: v_top² = v₀² - 4gl ≥ gl. Energy conservation + tension at any angle.

Multi StepHard

Common distractors

uses only energy not tension

Energy gives speed but tension constraint sets the floor

Sources

NCERT refs: Class 11 Physics Chapter 5, p.3

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Work Energy Theorem

Lesson

Practice MCQs

Quick recall before you leave

Work-energy theorem

Work by a constant force

Work by a variable force

Worked Example

Before solving, remember these

Formulas

Valid when

Do NOT use when

Valid when

Do NOT use when

Valid when

Valid when

Do NOT use when

Valid when

Valid when

Valid when

Do NOT use when

Valid when

Do NOT use when

Valid when

Valid when

Exam Traps & Common Mistakes

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

Correction

Wrong option pattern

Correction

Wrong option pattern

Correction

Wrong option pattern

Correction

Wrong option pattern

Past Year Questions

How NEET usually asks this

Common distractors

Common distractors

Common distractors

Common distractors

Common distractors

Common distractors

Common distractors

Sources

Free NEET study resources